A) \[10\]
B) \[12\]
C) -1
D) 2
Correct Answer: A
Solution :
Given series is \[3,\,\,7,\,\,11,\,\,15,...\]whose first term is\[3\], common difference \[4\] and sum is \[210\]. Let n be the number of term. \[\therefore \] \[S=\frac{n}{2}[2a+(n-1)d]\] \[\Rightarrow \] \[210=\frac{n}{2}[6+(n-1)4]\] \[\Rightarrow \] \[210=n[3+2n-2]\] \[\Rightarrow \] \[2{{n}^{2}}+n-210=0\] \[\Rightarrow \] \[n=10\]You need to login to perform this action.
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